September 19, 2020, 09:54:18 pm

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#### SmartWorker

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##### Re: Specialist 1/2 Question Thread!
« Reply #345 on: August 04, 2020, 01:18:53 pm »
0
Hey,

How would you do this question? (attached it below). I can do a and c once I have looked at a graph of it on desmos (attached below). But I would have not know how to do it otherwise.

What is the transformations that takes f(x) = arccos (x) to f(x) = arccos (6/x) (I was thinking maybe its a dilation by a factor of x^2/6 from the y-axis - but idk how this works?)

Thank you!
« Last Edit: August 04, 2020, 01:20:41 pm by SmartWorker »
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#### keltingmeith

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##### Re: Specialist 1/2 Question Thread!
« Reply #346 on: August 04, 2020, 01:34:09 pm »
+3
Hey,

How would you do this question? (attached it below). I can do a and c once I have looked at a graph of it on desmos (attached below). But I would have not know how to do it otherwise.

Super important - there's nothing wrong with this. Don't forget that you have your calculator available to you in about 2/3 of your exam time, so don't ever feel bad about the quicker method being by calculator - if not your only method. Tbh, I think this question would be quite mean to ask in non-calc exam, but definitely doable.

We know that the domain for arcos(x) is [-1,1]. This means that for any function f(x), arcos(f(x)) only works when -1<=f(x)<=1. So, if f(x) is 6/x, then we know that we MUST have:

$-1\leq \frac{6}{x}\leq 1$

Alright - so here's where things get tricky. If we want to make x the numerator, we can't do this in our current form - we need to do this by splitting up the inequality. So, we need to satisfy the following two equations:

$-1\leq\frac{6}{x}\text{ AND } \frac{6}{x}\leq 1$

Do you think you'd be able to take over from here?

b should be straightforward, it's just plugging into the differentiation rule for arcos(f(x)). Sketching the graph is straight up outside the study design, I wouldn't worry about it (unless someone wants to show me otherwise, but I don't remember ever seeing it)

What is the transformations that takes f(x) = arccos (x) to f(x) = arccos (6/x) (I was thinking maybe its a dilation by a factor of x^2/6 from the y-axis - but idk how this works?)

Thank you!

AFAIK, what you've said is right - I don't think the typical DRT transformation scheme really has a point for reciprocal or composition of functions.

In fact, checking the study design AGAIN, it only mentions linear transformations - this function is not a linear transformation of arcos(x) OR of 1/x, so you don't need to worry about specifying the transformations of this type.
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#### schoolstudent115

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##### Re: Specialist 1/2 Question Thread!
« Reply #347 on: August 04, 2020, 01:49:02 pm »
+3
1. Domain of arccos(x) = [-1,1]
Domain of Arccos(6/x): $6/x \in [-1,1]$
If you imagine a graph of $y=\frac{6}{x}$ you will realise that this means that: $x: x\in (-\infty,-6] \cup [6,\infty)$.

2. F’(x): use chain rule and distributing any powers along the way ~
$\frac{dy}{dx} = \frac{1}{x^2 * \sqrt{1-\frac{36}{x^2}}}$ .  Now that fraction is only positive if the denominator is positive:
$x^2 * \sqrt{1-\frac{36}{x^2}} > 0$

Well x^2>0 for all x except 0.

So this means that for the whole thing to be positive, then $\sqrt{1-\frac{36}{x^2}} > 0$ (+ * + == +).

Rearranging and simplifying: $x>6$.

So now we have two intersecting conditions: x is any number except for 0, and x>6. Intersecting the sets and this means that x>6. Done.

3. As for the graph, we know 3 things:
- It must be strictly increasing over x>6, but it increases slower as x—>infty (the derivative’s denominator will grow due to the x^2 term)
- we know the domain from (1)
- we know that on the LHS of the domain (-inf,-6] is mapped from (-1,0] (from the original domain of arccos(x)). And on arccos(x), over x: (-1,0] it is decreasing, so since we are using 6/x, then over (-inf,-6) it is increasing, or likewise if we flip the domain (-6,-inf) it is decreasing (from right to left).

This is enough info to graph.

But I’m not sure that is precisely how they’d do it.
« Last Edit: August 04, 2020, 01:56:13 pm by schoolstudent115 »
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#### S_R_K

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##### Re: Specialist 1/2 Question Thread!
« Reply #348 on: August 04, 2020, 02:53:15 pm »
+1
Note that arccos(1/x) = arcsec(x), so you can graph arccos(1/x) by graphing sec(x) for 0 ≤ x ≤ pi, and then reflecting about y = x.

#### schoolstudent115

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##### Re: Specialist 1/2 Question Thread!
« Reply #349 on: August 04, 2020, 03:30:15 pm »
+1
F=Arccos(1/x)
G=arcsec(x)
X = sec(G)=1/cos(G)
1/X = cos(F)
X = 1/cos(F) hence cos(F)=cos(G) ==~== F =~= G.

I didn’t think of it like that.
I’d graph cos(x)—>sec(x)—>arcsec(x)—>~—>arccos(1/x)—>dilate
The relation didn’t seem so obvious as the 1/x was wrapped up in the function.
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#### S_R_K

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##### Re: Specialist 1/2 Question Thread!
« Reply #350 on: August 04, 2020, 03:56:34 pm »
+1
F=Arccos(1/x)
G=arcsec(x)
X = sec(G)=1/cos(G)
1/X = cos(F)
X = 1/cos(F) hence cos(F)=cos(G) ==~== F =~= G.

In general this reasoning is a bit suspicious. It only works if you make some assumptions about F and G being one-to-one functions of x. (Which is true in this case, but be careful in general).

#### aspiringantelope

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##### Re: Specialist 1/2 Question Thread!
« Reply #351 on: August 08, 2020, 12:56:01 pm »
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Hello,
Do Trig Sum/Difference formulae work for sin^2,cos^2 and tan^2?
Like does this work?

$sin^2\left(x\right)cos^2\left(a\right)-cos^2\left(x\right)sin^2\left(a\right)$ = sin^2(x-a)

Thank you!
« Last Edit: August 08, 2020, 12:59:16 pm by aspiringantelope »

#### S_R_K

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##### Re: Specialist 1/2 Question Thread!
« Reply #352 on: August 08, 2020, 02:50:44 pm »
+3
Hello,
Do Trig Sum/Difference formulae work for sin^2,cos^2 and tan^2?
Like does this work?

$sin^2\left(x\right)cos^2\left(a\right)-cos^2\left(x\right)sin^2\left(a\right)$ = sin^2(x-a)

Thank you!

No. $\sin^2(u-v) = (\sin(u)\cos(v)-\sin(v)\cos(u))^2 \neq (\sin(u)\cos(v))^2-(\sin(v)\cos(u))^2$

The equation you wrote above is just the common howler $(x - a)^2 = x^2 - a^2$.

#### redset8

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##### Re: Specialist 1/2 Question Thread!
« Reply #353 on: August 11, 2020, 04:54:24 pm »
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Hi,
Encountered this question, but don't know how to go about proving it. If anyone could help, that would be much appreciated!
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#### keltingmeith

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##### Re: Specialist 1/2 Question Thread!
« Reply #354 on: August 11, 2020, 06:09:32 pm »
+5
Hi,
Encountered this question, but don't know how to go about proving it. If anyone could help, that would be much appreciated!

Oh boy, love a good circle proof. Everything I'm spitting out is literally me working through this question in real time - that means at some point I may hit a block or have to disregard something entirely. I think it's important to SHOW this process, because a lot of students will often ask, "but how did you know to try x?" - the true, honest to goodness, answer is - I didn't know to try x, I just did it, and it happened to work out.

Okay, so the question is - how can we prove this? Well, we know that all tangents to a circle meet the radius at 90 degrees - so maybe we can prove that each PQ and PR are parrallel to the respective tangents at the points PQ and PR cross on the circle edge. BUT, that would only work if P is the middle of the circle - which isn't necessarily true.

Okay, instead of questioning how to answer the question, let's instead just figure out what we know and work from there. So, <ADC and <ABC must add to 180 degrees, and the same for <DAB and <DCB. But, I'm not sure how we could use this... But it does mean we know the angles <QBC and <RDC. It's worth noting down, even if we never use it. Let's say <ADC=<QBC=d, and <ABC=<RDC=b.

In fact, we can use this to figure out all the angles in the triangle. Using the same logic, <RCD=<QCD=<DAB=a (again, I've just decided that it should be called a, because that makes things easier). So that means that we now know all the angles in the two triangles they made up, RDC and QBC. In fact, now we know:

<DRC=180-b-a
<BQC=180-d-a

So, how does that help us? Well, let's call the intersection between QD and RP, M, and the intersection between RB and QP, N (I hope you're labelling this on your own circle), then all we need to prove is that either <MQP + <QMP = 90 degrees OR <NRP + <RNP = 90 degrees. And, using the above, we know that:

<DRC = 2*<NRP = 180 - b - a, <NRP = 90 - (b+a)/2
<BQC = 2*<MQP = 180 - d - a, <MQP = 90 - (d+a)/2

So for simplicity, I'm just going to work on <PNR. So I know for <NRP + <PNR = 90, then I need <PNR = (b+a)/2 - the question is, how can I get there? Well, I know that <PNR=<BNQ=180-<CNQ=180-<PNB. Can I figure any of those out?

The problem with any of these triangles is that N doesn't lie on the circle - it's inside the circle, and we don't have many proofs that rely on working INSIDE the circle - so frustratingly, knowing we're so close, this approach looks doomed to fail.

Okay, new idea. What if we made a circle using the points P, Q, and R. Then, all we'd need to do is prove that the length QR passes through the centre of that circle. Then the angle <QPR HAS to be 90 degrees.

... Except I'm not sure how we could use any of the information we've been given to inform ourselves of this circle we just made. Okay, maybe I'll go back to the (a+b)/2 thing

Okay, so let's try a slightly different approach, and we're going to extend QP so that it intersects with AR - and we'll call that point Z because I'm running out of letters that make sense. Since RP bisects the triangle ZRN, if PQ and PR are perpendicular, then <RZN=<RNZ. No, I'm stopping this train of thought, things are just going to get worse before they get better. There's gotta be a simpler way.

Okay, I think I've got it. If I go back to the (a+b)/2 thing, I know that <QNB=<RNP, and I want to show that <RNP=(a+b)/2. I may have realised this earlier if I labelled myself (oops), but I just realised that I CAN figure out <QNB because I know all the other angles in the triangle QNB:

<BQN=<MQP=90-(d+a)/2
<NBQ=d
<QNB=180-<BQN-<NBQ=180-(90-d/2-a/2)-d=90+a/2-d/2=90+(a-d)/2

Okay, not looking good. But, we know that d=180-b, so we then get:

<RNP=<QNB=90+a/2 - (180-b)/2=90 + a/2 - 180/2 + b/2 = 90 + a/2 - 90 + b/2 = (a+b)/2

Which means that:

<RPQ=<RPN=180-<RNP-<NRP=180-(a+b)/2-(90-(a+b)/2)=90

Fuck, this was a work-out. I'm sure this got confusing along the way, so try re-doing this yourself to see if it makes more sense

EDIT: After giving up I hit a break-through, so hopefully you've noticed this and come back after I made my initial post? Oops
« Last Edit: August 11, 2020, 06:27:05 pm by keltingmeith »
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#### redset8

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##### Re: Specialist 1/2 Question Thread!
« Reply #355 on: August 11, 2020, 08:57:30 pm »
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Wow. Just wow. I started out like you, thinking about cyclic quadrilaterals, and then couldn't really progress.
I've worked through your solution in my book. I'll now try it again by myself.

Which means that:

<RPQ=<RPN=180-<RNP-<NRP=180-(a+b)/2-(90-(a+b)/2)=90

Just the very last point (I was so close...), how does <NRP = (90-(a+b)/2). Maybe I'm just not seeing it tonight...
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#### keltingmeith

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##### Re: Specialist 1/2 Question Thread!
« Reply #356 on: August 12, 2020, 08:34:22 am »
+2
Just the very last point (I was so close...), how does <NRP = (90-(a+b)/2). Maybe I'm just not seeing it tonight...

So that was something I proved much earlier in the process. Remember that the line BP bisects the angle <BRA, and so the angles <DRP and <PRC (which is the same as <PRN, since R, N, and C are collinear) are exactly half the angle <BRA. If the question is what's the angle <BRA - since we decided that the angles <RAB=a, and <ABR=b, then angle <BRA = 180-a-b. Cut that in half, and you get <NRP
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#### redset8

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##### Re: Specialist 1/2 Question Thread!
« Reply #357 on: August 13, 2020, 04:48:04 pm »
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Ah, thanks heaps! Geometry practice for me...
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##### Re: Specialist 1/2 Question Thread!
« Reply #358 on: August 13, 2020, 05:19:16 pm »
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Can someone please show me step by step how I'd solve this? Thanks

#### keltingmeith

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##### Re: Specialist 1/2 Question Thread!
« Reply #359 on: August 13, 2020, 05:24:05 pm »
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Can someone please show me step by step how I'd solve this? Thanks

What are you struggling with? What have you tried yourself?
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